Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <string.h>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#include <random>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define df(x) int x = in()
#define dame { puts("-1"); return 0;}
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,vector<T>,greater<T> >
#define bn(x) ((1<<x)-1)
#define newline puts("")
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}
template<typename T>istream& operator>>(istream&i,const vector<T>&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const vector<T>&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,const vector<T>&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,const pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const int MX = 100005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;

// Segment tree (RMQ type)
struct seg {
  typedef int TT;
  vector<TT> d; int x2;
  seg(){}
  seg(int mx) {
    x2 = 1; while (x2 < mx) x2 <<= 1;
    d.resize(x2<<1,-1);
  }
  TT get(int i) {
    TT res = -1;
    i += x2;
    for (;i;i>>=1) {
      maxs(res,d[i]);
    }
    return res;
  }
  void add(int a, int b, TT x) { return add(a,b,x,1,0,x2);}
  void add(int a, int b, TT x, int i, int l, int r){
    if (a <= l && r <= b) {
      d[i] = x;
      return;
    }
    int c = (l+r)>>1;
    if (a < c) add(a,b,x,i<<1,l,c);
    if (c < b) add(a,b,x,i<<1|1,c,r);
  }
};
//

// coordinate compression
struct X {
  typedef int T;
  vector<T> d;
  void add(T x) { d.pb(x);}
  void init() {
    sort(rng(d));
    d.erase(unique(rng(d)), d.end());
  }
  int size() const { return sz(d);}
  int operator[](int i) const { return d[i];}
  int operator()(T x) const { return upper_bound(rng(d),x)-d.begin()-1;}
};
//



int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  vi a(n), b(n);
  rep(i,n) {
    scanf("%d%d",&a[i],&b[i]);
    a[i] *= 2;
    --b[i];
  }
  ll ans = LINF, sum = 0;
  vl s(n);
  X xs;
  xs.add(0);
  rep(i,n) {
    s[i] = sum+1;
    sum += a[i];
    xs.add(s[i]%m);
    xs.add(sum%m);
  }
  xs.init();
  seg t(sz(xs));
  vl d(n);
  rep(i,n) {
    if (b[i]) continue;
    if (a[i] > m) dame;
    int l = xs(s[i]%m);
    int r = xs((s[i]+a[i]-1)%m);
    int p = t.get(r);
    d[i] = 0;
    if (p != -1) {
      ll r1 = (s[i]+a[i]-1)%m;
      ll r2 = (s[p]+a[p]-1)%m;
      d[i] = d[p]+(r2+m-r1)%m;
    }
    if (l <= r) {
      t.add(l,r,i);
    } else {
      t.add(l,sz(xs),i);
      t.add(0,r,i);
    }
  }
  rep(i,sz(xs)) {
    int x = xs[i];
    int p = t.get(i);
    if (p == -1) {
      ans = 0;
      break;
    }
    ll r2 = (s[p]+a[p]-1)%m;
    ll now = d[p]+(r2+m-x)%m;
    // cout<<i<<" "<<x<<" "<<now<<endl;
    mins(ans,now);
  }
  ans += sum;
  cout<<ans<<endl;
  return 0;
}